Question

Balance the following redox reactions by the ion-electron method in the basic medium.
${ MnO }_{ 4 }^{ - }{ (aq)+{ I }^{ - }{ (aq)\rightarrow { MnO }_{ 2 }{ (s)+{ I }_{ 2 }{ (s) } } } }$

Answer 1

Hint: Ion-electron method: It is also known as the half-reaction method, the redox reaction is isolated into two-half conditions: one for reduction and one for oxidation. Each one of these half-reactions is balanced independently and then combined to give a balanced redox reaction.

Complete step by step solution:
There are following steps to balance a redox reaction by the ion-electron method in a basic medium which is explained below:
Step 1: The two half-reactions intricate in the given reaction are:
Oxidation: ${ I }_{ 2 }{ (aq)\rightarrow I }_{ 2 }{ (s) }$
Reduction: ${ MnO }_{ 4 }^{ 2- }{ (aq)\rightarrow MnO }_{ 2 }{ (aq) }$

Step 2: Balance I in the oxidation half-reaction, we have
${ 2I }^{ - }{ (aq)\rightarrow I }_{ 2 }{ (s) }$
Now, to balance the charge we have to add two electrons in RHS of the reaction.
${ 2I }^{ - }{ (aq)\rightarrow I }_{ 2 }{ (s)+2e^{ - } }$

Step 3:The oxidation state of Mn has reduced from +7 to +4 in the reduction half-reaction.
Thus, three electrons are added to the LHS of the reaction.
${ MnO }_{ 4 }^{ - }{ (aq)+{ 3e }^{ - } }{ \rightarrow MnO }_{ 2 }{ (s) }$
Now, to balance the charge we have to add four ions to the RHS of the reaction as the reaction is taking place in a basic medium.
${ MnO }_{ 4 }^{ - }{ (aq)+{ 3e }^{ - } }{ \rightarrow MnO }_{ 2 }{ (s)+4OH }^{ - }$

Step 4: In this reaction, there are six O atoms on the RHS and four O atoms on the LHS. Therefore, two water molecules are added to the LHS.
${ MnO }_{ 4 }^{ - }{ (aq)+{ 2H }_{ 2 }{ O+3e }^{ - } }{ \rightarrow MnO }_{ 2 }{ (s)+4OH }^{ - }$

Step 5: Now, equalize the number of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we have:
${ 6I }^{ - }{ (aq)\rightarrow 3I }_{ 2 }{ (s)+2e^{ - } }$
${ 2MnO }_{ 4 }^{ - }{ (aq)+{ 4H }_{ 2 }{ O+6e }^{ - } }{ \rightarrow 2MnO }_{ 2 }{ (s)+8OH }^{ - }$

Step 6: Now, by adding the two half-reactions, we have the net balanced redox reaction as:
${ 6I }^{ - }{ +2MnO }_{ 4 }^{ - }{ (aq)+{ 4H }_{ 2 } }{ O\rightarrow 2MnO }_{ 2 }{ (s) }{ +3I }_{ 2 }{ (s)+8OH^{ - } }$

Note: The possibility to make a mistake is to balance a charge in oxidation half-reactions by adding electrons in the product, not a reactant, and in reduction half-reaction, the charge is balanced by adding electrons as a reactant.